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From: Jon Watte <hplus@mi...>  20050729 16:42:44

> the "log3 rule" IF you manually introduce panning into your mix. By that, I meant the "3 dB pan rule"; sorry about the typo. Cheers, / h+   The early bird gets the worm, but the second mouse gets the cheese. 
From: Jon Watte <hplus@mi...>  20050729 16:36:49

The "official" mixing formula is to just add them. The log formulas have to do with converting between voltage values (what samples are) and decibels (which you can show to the user) and will sound like crap if you play them. The average formula is just the same as adding, and then reducing gain by 6 dB to avoid clipping. (A+B)/2 However, you're talking about two speakers, without any more qualification. There is a difference in perceived volume between a signal playing out just a single speaker, and a signal playing out two speakers. Typically, you'll compensate for this using something called the "log3 rule" IF you manually introduce panning into your mix. However, for cases like a stereo>mono downmix, it's just mixing, and you just add the numbers. Cheers, / h+ Gaël LEQUEUX wrote: > Hi there, > I'm currently searching an accurate sound mixing algorithm. > I have two 16 Bits samples S1 and S2 and I'd like to find SMix, the new > sample I must put in my sound data to obtain the exact same volume as S1 and > S2 were played on 2 speakers. > > The formulas I found are (Of course I have to saturate SMix after that): >  Simply add S1 and S2 >  Average S1 and S2 >  SMix = Log(10^S1+10^S2) >  SMix = 20*Log(10^(S1/20)+10^(S2/20)) > > Is there an 'official' formula? > > Cheers, > Gaël LEQUEUX > > > >  > SF.Net email is Sponsored by the Better Software Conference & EXPO September > 1922, 2005 * San Francisco, CA * Development Lifecycle Practices > Agile & PlanDriven Development * Managing Projects & Teams * Testing & QA > Security * Process Improvement & Measurement * http://www.sqe.com/bsce5sf > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_ida88 > >   The early bird gets the worm, but the second mouse gets the cheese. 
From: Nils Pipenbrinck <np@in...>  20050729 15:41:51

Gaël LEQUEUX wrote: >Hi there, >I'm currently searching an accurate sound mixing algorithm. >I have two 16 Bits samples S1 and S2 and I'd like to find SMix, the new >sample I must put in my sound data to obtain the exact same volume as S1 and >S2 were played on 2 speakers. > > Hi Gaël, You'll simply want to add the samples. There's always a lot of confusion when it comes to energy, multiple speakers and perceived volume. Playing the same signal twice on two loudspeakes doubles the energy but not the perceived volume. Adding those samples and playing them over one speaker does the same (assuming that the speakers and amplifiers work linear, which I do). You need the logequations when you want to double the perceived volume or you want to mix two signals and have the same perceived volume as a single signal. Nils (and now I'm expecting to get yet another autoreply from Dennis :) 
From: Dennis Gustafsson <dennis@me...>  20050729 15:24:29

Hi, I am currently on vacation and will be back on August 22. For business inqueries please contact jonas@... For technical inqueries, contact support@... Regards, Dennis 
From: <gael.sc2x@la...>  20050729 15:23:45

Hi there, I'm currently searching an accurate sound mixing algorithm. I have two 16 Bits samples S1 and S2 and I'd like to find SMix, the new sample I must put in my sound data to obtain the exact same volume as S1 = and S2 were played on 2 speakers. The formulas I found are (Of course I have to saturate SMix after that):  Simply add S1 and S2  Average S1 and S2  SMix =3D Log(10^S1+10^S2)  SMix =3D 20*Log(10^(S1/20)+10^(S2/20)) Is there an 'official' formula? Cheers, Ga=EBl LEQUEUX 
From: <gdalgo@er...>  20050722 17:16:21

>Traffic has seemed awfully light recently on gdalgo... in case you are bored and have windows machine, watch these demos. http://www.continuousphysics.com/p/bullet0.1demos.zip Also you can visit my continuous collision detection and physics forum, for the source, demos and discussions in that area: http://www.continuousphysics.com/Bullet/phpBB2/index.php [work in progress] Erwin Coumans 
From: Andras Balogh <andras.balogh@gm...>  20050722 14:47:22

All the mails were sent to this list, looks like you have a problem with your subscription/mail account... I'll forward you those messages in private. cheers, Andras Bill Baxter wrote: > Were those private messages? All I could see of that conversation > with Mick was the snippet you quoted which didn't make any sense out > of context. > > If it was a personal conversation, would you mind repeating what the > solution was? > > If it took place on the list here, then seems I've got a problem with > the list. Traffic has seemed awfully light recently on gdalgo... > > bb 
From: Willem de Boer <wdeboer@pl...>  20050722 06:53:42

For the bored, a sort of handwavy geometric proof=20 of this would go like this: For a rectangle, R, with edges {e1,e2,e3,e4}, and normal n. Edge e1 being fixed, rotate R about e1=20 such that n and the view vector v lie on the same line, call this normal n_m. Call e3 the edge opposite to e1. Let the screenspace area of R=20 be a function of its normal, A(n). We want to show that n_m maximises A(n). Let V be the plane constructed from the viewpoint and e3. Construct a cylinder C with its axis parallel to e1 and such that e1 and its opposite edge, call it e3, lie on the surface of the cylinder. Then the surface of the cylinder represents the constraint of e3, when we rotate R about e1. Also V is a tangent plane of the cylinder. We now note=20 that in order for A(n) to not be a maximum at n_m,=20 the cylinder would have to intersect V somewhere else than e3(n_m), but it doesn't because it is=20 a tangent plane to C. Hmmm, very handwavy, but by its geometric construction intuitively understandable.  Willem H. de Boer Homepage: http://www.whdeboer.com=20 > Original Message > From: gdalgorithmslistadmin@...=20 > [mailto:gdalgorithmslistadmin@...] On=20 > Behalf Of Megan Fox > Sent: Thursday, July 21, 2005 9:32 PM > To: gdalgorithmslist@... > Subject: Re: [Algorithms] rotating in world space to have=20 > result in screen space >=20 > Forgive me if this is off, it's one of those "shouldn't this work?" > sorts of intuition suggestions: >=20 > The point at which a quad is most visible is always going to=20 > be when the normal of that quad is pointed directly at the=20 > viewer's position (or directly away from, but then it gets=20 > discarded, so ignore that).=20 > So, isn't the real question the closest you can rotate the=20 > normal to "camera lookat"? >=20 >=20 > Here's where it gets even more dicey... shouldn't=20 > transforming the normalized quadtocamera vector into the=20 > plane you're describing be able to give you the data=20 > necessary to generate the optimal transform? >=20 >=20 > Sorry, I have no math to prove this, it just seems as if this "should" > work? ... if I have time later, I'll try and work out an=20 > equation for all of this. >=20 > Megan Fox >=20 > On 7/21/05, Andras Balogh <andras.balogh@...> wrote: > > Ok, so here's the original problem: > >=20 > > I have a rectangle in world space, two of its neighbouring vertices=20 > > are locked at a fixed position (so you can rotate the=20 > rectangle around=20 > > the locked edge). Now, given an arbitrary viewing transform, rotate=20 > > the rectangle, so that its screen space size is maximized! > >=20 > > I was thinking that I'll have the maximum area, if the=20 > normal vector=20 > > of the rotated rectangle is parallel in screen space to the=20 > projected=20 > > locked edge. So, having the screen space normal, I could=20 > compute the=20 > > angle between that and the screen space up vector. If I could=20 > > transform this angle back to world space, then I could just=20 > rotate the=20 > > rectangle by this amout.. > >=20 > > Ehh, I hope it makes some sense, there more I think about=20 > it, the more=20 > > confused I become :) > >=20 > >=20 > > Andras > >=20 > >=20 > > Ralph Egas wrote: > > > Hi Andras, > > > > > > Are you looking for a solution for an arbitrary 'r' and=20 > 'p'? What if 'r' > > > gets perpendicular to the right or up vector in screenspace?=20 > > > Obviously then there's not going to be any solution for=20 > any world space angle.. > > > > > > If you could just elaborate your idea and what you need this=20 > > > solution for, maybe that could help. > > > > > > Cheers, > > > > > > Ralph > > > > > >  Original Message  From: "Andras Balogh"=20 > > > <andras.balogh@...> > > > To: <gdalgorithmslist@...> > > > Sent: Thursday, July 21, 2005 8:02 PM > > > Subject: [Algorithms] rotating in world space to have result in=20 > > > screen space > > > > > > > > >> I have a point 'v0' in world space, that I can only=20 > rotate around a=20 > > >> fixed point 'p' and axis 'r'. So movement is constrained=20 > on a circle. > > >> Now, I'm looking for the point 'v1' on this circle, so that the=20 > > >> three points (v0, p, v1), when projected to the screen,=20 > enclose a given angle. > > >> > > >> Any ideas how to compute the world space 'v1', given the screen=20 > > >> space angle? > > >> > > >> I was thinking of working backwards, rotating first in screen=20 > > >> space, using the given angle, and then trying to figure out the=20 > > >> world space position of v1, but I'm a bit afraid, that this=20 > > >> wouldn't be a very robust solution.. > > >> > > >> thanks, > > >> > > >> > > >> Andras > >=20 > >=20 > >  > > SF.Net email is sponsored by: Discover Easy Linux Migration=20 > Strategies=20 > > from IBM. Find simple to follow Roadmaps, straightforward articles,=20 > > informative Webcasts and more! Get everything you need to get up to=20 > > speed, fast. = http://ads.osdn.com/?ad_id=3D7477&alloc_id=3D16492&op=3Dclick > > _______________________________________________ > > GDAlgorithmslist mailing list > > GDAlgorithmslist@... > > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > > Archives: > > http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 > > >=20 >=20 >  > SF.Net email is sponsored by: Discover Easy Linux Migration=20 > Strategies from IBM. Find simple to follow Roadmaps,=20 > straightforward articles, informative Webcasts and more! Get=20 > everything you need to get up to speed, fast.=20 > http://ads.osdn.com/?ad_idt77&alloc_id=16492&op=3Dick > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_ida88 >=20 
From: Bill Baxter <wbaxter@gm...>  20050722 03:07:05

Were those private messages? All I could see of that conversation with Mick was the snippet you quoted which didn't make any sense out of context. If it was a personal conversation, would you mind repeating what the solution was? If it took place on the list here, then seems I've got a problem with the list. Traffic has seemed awfully light recently on gdalgo... bb On 7/22/05, Andras Balogh <andras.balogh@...> wrote: > Hey Bill, I've just finished implementing the solution suggested by Mick > and Megan, and it works like a charm, and result looks great!! And this > solution is also very simple and super robust! So I think I'll stick to > it! :) >=20 > Thanks, >=20 > Andras >=20 >=20 > Bill Baxter wrote: > > Seems like a small extension of standard raypicking would do the trick= . > > You can get one world space ray from the camera location and world > > space point p . > > Get another world space ray by taking some point on the screen with > > the desired angle and turning that into a ray using the standard way > > used raypicking techniques. > > > > Those two world space rays define a world space plane (they both > > contain the camera origin, so it is a plane.) > > > > Now solve a quadratic equation to compute the intersections of the > > world space circle with the world space plane. Pick the one of the > > two answers that's right. > > > > > > bb > > > > On 7/22/05, Andras Balogh <andras.balogh@...> wrote: > > > >>I have a point 'v0' in world space, that I can only rotate around a > >>fixed point 'p' and axis 'r'. So movement is constrained on a circle. > >>Now, I'm looking for the point 'v1' on this circle, so that the three > >>points (v0, p, v1), when projected to the screen, enclose a given angle= . > >> > >>Any ideas how to compute the world space 'v1', given the screen space a= ngle? > >> > >>I was thinking of working backwards, rotating first in screen space, > >>using the given angle, and then trying to figure out the world space > >>position of v1, but I'm a bit afraid, that this wouldn't be a very > >>robust solution.. > >> > >>thanks, > >> > >> > >>Andras > >> > >> > >> > >>SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > >>from IBM. Find simple to follow Roadmaps, straightforward articles, > >>informative Webcasts and more! Get everything you need to get up to > >>speed, fast. http://ads.osdn.com/?ad_id=3D7477&alloc_id=3D16492&op=3Dcl= ick > >>_______________________________________________ > >>GDAlgorithmslist mailing list > >>GDAlgorithmslist@... > >>https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > >>Archives: > >>http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 > >> > > > > > > >=20 >=20 >=20 >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=3D7477&alloc_id=3D16492&op=3Dclic= k > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 >=20 =20 William V. Baxter III, Ph.D. OLM Digital Kono Dens Building Rm 302 188 Wakabayashi Setagayaku Tokyo, Japan 1540023 +81 (3) 54335588 
From: Andras Balogh <andras.balogh@gm...>  20050722 00:52:44

Hey Bill, I've just finished implementing the solution suggested by Mick and Megan, and it works like a charm, and result looks great!! And this solution is also very simple and super robust! So I think I'll stick to it! :) Thanks, Andras Bill Baxter wrote: > Seems like a small extension of standard raypicking would do the trick. > You can get one world space ray from the camera location and world > space point p . > Get another world space ray by taking some point on the screen with > the desired angle and turning that into a ray using the standard way > used raypicking techniques. > > Those two world space rays define a world space plane (they both > contain the camera origin, so it is a plane.) > > Now solve a quadratic equation to compute the intersections of the > world space circle with the world space plane. Pick the one of the > two answers that's right. > > > bb > > On 7/22/05, Andras Balogh <andras.balogh@...> wrote: > >>I have a point 'v0' in world space, that I can only rotate around a >>fixed point 'p' and axis 'r'. So movement is constrained on a circle. >>Now, I'm looking for the point 'v1' on this circle, so that the three >>points (v0, p, v1), when projected to the screen, enclose a given angle. >> >>Any ideas how to compute the world space 'v1', given the screen space angle? >> >>I was thinking of working backwards, rotating first in screen space, >>using the given angle, and then trying to figure out the world space >>position of v1, but I'm a bit afraid, that this wouldn't be a very >>robust solution.. >> >>thanks, >> >> >>Andras >> >> >> >>SF.Net email is sponsored by: Discover Easy Linux Migration Strategies >>from IBM. Find simple to follow Roadmaps, straightforward articles, >>informative Webcasts and more! Get everything you need to get up to >>speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click >>_______________________________________________ >>GDAlgorithmslist mailing list >>GDAlgorithmslist@... >>https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist >>Archives: >>http://sourceforge.net/mailarchive/forum.php?forum_id=6188 >> > > > 
From: Bill Baxter <wbaxter@gm...>  20050722 00:22:55

Seems like a small extension of standard raypicking would do the trick. You can get one world space ray from the camera location and world space point p . Get another world space ray by taking some point on the screen with the desired angle and turning that into a ray using the standard way used raypicking techniques. Those two world space rays define a world space plane (they both contain the camera origin, so it is a plane.) Now solve a quadratic equation to compute the intersections of the world space circle with the world space plane. Pick the one of the two answers that's right. bb On 7/22/05, Andras Balogh <andras.balogh@...> wrote: > I have a point 'v0' in world space, that I can only rotate around a > fixed point 'p' and axis 'r'. So movement is constrained on a circle. > Now, I'm looking for the point 'v1' on this circle, so that the three > points (v0, p, v1), when projected to the screen, enclose a given angle. >=20 > Any ideas how to compute the world space 'v1', given the screen space ang= le? >=20 > I was thinking of working backwards, rotating first in screen space, > using the given angle, and then trying to figure out the world space > position of v1, but I'm a bit afraid, that this wouldn't be a very > robust solution.. >=20 > thanks, >=20 >=20 > Andras >=20 >=20 >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=3D7477&alloc_id=3D16492&op=3Dclic= k > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 >=20 =20 William V. Baxter III, Ph.D. OLM Digital Kono Dens Building Rm 302 188 Wakabayashi Setagayaku Tokyo, Japan 1540023 +81 (3) 54335588 
From: Andras Balogh <andras.balogh@gm...>  20050721 20:58:02

Mick West wrote: > Andras Balogh wrote: > >> Mick West wrote: >> >>> Actualy, thinking about it a bit more, that only minimizes the angle >>> between the rectangles normal and the viewer. I think this >>> appoximates maximizing the screen area, but might be especially wrong >>> when very close to the camera. >> >> Why is that? I can't see why minimizing the normal/view vector angle >> would be wrong, could you give an example? >> > Hold a book about a foot in front of your face, close one eye, tilt the > top edge towards you. I'm not at all sure, but you could imagine the > widening of the projection of the book due to perspective might be more > (in terms of area) than the vertical foreshortening, especially with a > very wide angle lens. Still, it should be fine for your application. Hah, that's interesting! Fortunately perspective not only adds to the area on the near side, but also removes on the far side! Anyway, I think I can safely ignore that! :) Thanks again, Andras 
From: Mick West <dev@mi...>  20050721 20:44:20

Andras Balogh wrote: > Mick West wrote: > >> Actualy, thinking about it a bit more, that only minimizes the angle >> between the rectangles normal and the viewer. I think this >> appoximates maximizing the screen area, but might be especially wrong >> when very close to the camera. > > > Why is that? I can't see why minimizing the normal/view vector angle > would be wrong, could you give an example? > Hold a book about a foot in front of your face, close one eye, tilt the top edge towards you. I'm not at all sure, but you could imagine the widening of the projection of the book due to perspective might be more (in terms of area) than the vertical foreshortening, especially with a very wide angle lens. Still, it should be fine for your application. Mick. 
From: Andras Balogh <andras.balogh@gm...>  20050721 20:34:12

Mick West wrote: > Actualy, thinking about it a bit more, that only minimizes the angle > between the rectangles normal and the viewer. I think this appoximates > maximizing the screen area, but might be especially wrong when very > close to the camera. Why is that? I can't see why minimizing the normal/view vector angle would be wrong, could you give an example? > What's the REAL problem? Why do you need the screen space maximized? Haha, that's the REAL problem! :) Really, I have to render text on these quads, and the more you see, the better. But the quads have to remain fixed at an edge (this is a design restriction). Cheers, Andras 
From: Andras Balogh <andras.balogh@gm...>  20050721 20:30:53

Megan Fox wrote: > Forgive me if this is off, it's one of those "shouldn't this work?" > sorts of intuition suggestions: > > The point at which a quad is most visible is always going to be when > the normal of that quad is pointed directly at the viewer's position > (or directly away from, but then it gets discarded, so ignore that). > So, isn't the real question the closest you can rotate the normal to > "camera lookat"? Hehe, I was just thinking the same during lunch, and I think Mick's solution is basically the same. Yeah, I think it should work, although I haven't verified it on paper myself yet. I'll do that now, and report back soon! :) Thanks, Andras 
From: Mick West <dev@mi...>  20050721 19:57:10

Actualy, thinking about it a bit more, that only minimizes the angle between the rectangles normal and the viewer. I think this appoximates maximizing the screen area, but might be especially wrong when very close to the camera. What's the REAL problem? Why do you need the screen space maximized? Mick. Mick West wrote: > If A and B are the fixed points, C is one of the other (movable) > points of the rectangle, and V is the viewpoint of the transform, P > is the plane at point A with normal AB and C1 is the projection of C > onto P then the area is maximized by rotating C1 into C2 where AC2 is > perpendicular to the plane defined by ABV. Rotate C (and D) by the > angle between C1 and C2 relative to A (or without C2, the angle > between A>C1 and the normal of ABV). > Mick. > > Andras Balogh wrote: > >> Ok, so here's the original problem: >> >> I have a rectangle in world space, two of its neighbouring vertices >> are locked at a fixed position (so you can rotate the rectangle >> around the locked edge). Now, given an arbitrary viewing transform, >> rotate the rectangle, so that its screen space size is maximized! >> >> I was thinking that I'll have the maximum area, if the normal vector >> of the rotated rectangle is parallel in screen space to the projected >> locked edge. So, having the screen space normal, I could compute the >> angle between that and the screen space up vector. If I could >> transform this angle back to world space, then I could just rotate >> the rectangle by this amout.. >> >> Ehh, I hope it makes some sense, there more I think about it, the >> more confused I become :) >> >> >> Andras >> >> >> Ralph Egas wrote: >> >>> Hi Andras, >>> >>> Are you looking for a solution for an arbitrary 'r' and 'p'? What if >>> 'r' gets perpendicular to the right or up vector in screenspace? >>> Obviously then there's not going to be any solution for any world >>> space angle.. >>> >>> If you could just elaborate your idea and what you need this >>> solution for, maybe that could help. >>> >>> Cheers, >>> >>> Ralph >>> >>>  Original Message  From: "Andras Balogh" >>> <andras.balogh@...> >>> To: <gdalgorithmslist@...> >>> Sent: Thursday, July 21, 2005 8:02 PM >>> Subject: [Algorithms] rotating in world space to have result in >>> screen space >>> >>> >>>> I have a point 'v0' in world space, that I can only rotate around a >>>> fixed point 'p' and axis 'r'. So movement is constrained on a >>>> circle. Now, I'm looking for the point 'v1' on this circle, so that >>>> the three points (v0, p, v1), when projected to the screen, enclose >>>> a given angle. >>>> >>>> Any ideas how to compute the world space 'v1', given the screen >>>> space angle? >>>> >>>> I was thinking of working backwards, rotating first in screen >>>> space, using the given angle, and then trying to figure out the >>>> world space position of v1, but I'm a bit afraid, that this >>>> wouldn't be a very robust solution.. >>>> >>>> thanks, >>>> >>>> >>>> Andras >>> >>> >> >> >>  >> SF.Net email is sponsored by: Discover Easy Linux Migration Strategies >> from IBM. Find simple to follow Roadmaps, straightforward articles, >> informative Webcasts and more! Get everything you need to get up to >> speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click >> _______________________________________________ >> GDAlgorithmslist mailing list >> GDAlgorithmslist@... >> https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist >> Archives: >> http://sourceforge.net/mailarchive/forum.php?forum_id=6188 >> > > > >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=6188 > 
From: Mick West <dev@mi...>  20050721 19:47:29

If A and B are the fixed points, C is one of the other (movable) points of the rectangle, and V is the viewpoint of the transform, P is the plane at point A with normal AB and C1 is the projection of C onto P then the area is maximized by rotating C1 into C2 where AC2 is perpendicular to the plane defined by ABV. Rotate C (and D) by the angle between C1 and C2 relative to A (or without C2, the angle between A>C1 and the normal of ABV). Mick. Andras Balogh wrote: > Ok, so here's the original problem: > > I have a rectangle in world space, two of its neighbouring vertices > are locked at a fixed position (so you can rotate the rectangle around > the locked edge). Now, given an arbitrary viewing transform, rotate > the rectangle, so that its screen space size is maximized! > > I was thinking that I'll have the maximum area, if the normal vector > of the rotated rectangle is parallel in screen space to the projected > locked edge. So, having the screen space normal, I could compute the > angle between that and the screen space up vector. If I could > transform this angle back to world space, then I could just rotate the > rectangle by this amout.. > > Ehh, I hope it makes some sense, there more I think about it, the more > confused I become :) > > > Andras > > > Ralph Egas wrote: > >> Hi Andras, >> >> Are you looking for a solution for an arbitrary 'r' and 'p'? What if >> 'r' gets perpendicular to the right or up vector in screenspace? >> Obviously then there's not going to be any solution for any world >> space angle.. >> >> If you could just elaborate your idea and what you need this solution >> for, maybe that could help. >> >> Cheers, >> >> Ralph >> >>  Original Message  From: "Andras Balogh" >> <andras.balogh@...> >> To: <gdalgorithmslist@...> >> Sent: Thursday, July 21, 2005 8:02 PM >> Subject: [Algorithms] rotating in world space to have result in >> screen space >> >> >>> I have a point 'v0' in world space, that I can only rotate around a >>> fixed point 'p' and axis 'r'. So movement is constrained on a >>> circle. Now, I'm looking for the point 'v1' on this circle, so that >>> the three points (v0, p, v1), when projected to the screen, enclose >>> a given angle. >>> >>> Any ideas how to compute the world space 'v1', given the screen >>> space angle? >>> >>> I was thinking of working backwards, rotating first in screen space, >>> using the given angle, and then trying to figure out the world space >>> position of v1, but I'm a bit afraid, that this wouldn't be a very >>> robust solution.. >>> >>> thanks, >>> >>> >>> Andras >> > > >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=6188 > 
From: Megan Fox <shalinor@gm...>  20050721 19:32:32

Forgive me if this is off, it's one of those "shouldn't this work?" sorts of intuition suggestions: The point at which a quad is most visible is always going to be when the normal of that quad is pointed directly at the viewer's position (or directly away from, but then it gets discarded, so ignore that).=20 So, isn't the real question the closest you can rotate the normal to "camera lookat"? Here's where it gets even more dicey... shouldn't transforming the normalized quadtocamera vector into the plane you're describing be able to give you the data necessary to generate the optimal transform? Sorry, I have no math to prove this, it just seems as if this "should" work? ... if I have time later, I'll try and work out an equation for all of this. Megan Fox On 7/21/05, Andras Balogh <andras.balogh@...> wrote: > Ok, so here's the original problem: >=20 > I have a rectangle in world space, two of its neighbouring vertices are > locked at a fixed position (so you can rotate the rectangle around the > locked edge). Now, given an arbitrary viewing transform, rotate the > rectangle, so that its screen space size is maximized! >=20 > I was thinking that I'll have the maximum area, if the normal vector of > the rotated rectangle is parallel in screen space to the projected > locked edge. So, having the screen space normal, I could compute the > angle between that and the screen space up vector. If I could transform > this angle back to world space, then I could just rotate the rectangle > by this amout.. >=20 > Ehh, I hope it makes some sense, there more I think about it, the more > confused I become :) >=20 >=20 > Andras >=20 >=20 > Ralph Egas wrote: > > Hi Andras, > > > > Are you looking for a solution for an arbitrary 'r' and 'p'? What if 'r= ' > > gets perpendicular to the right or up vector in screenspace? Obviously > > then there's not going to be any solution for any world space angle.. > > > > If you could just elaborate your idea and what you need this solution > > for, maybe that could help. > > > > Cheers, > > > > Ralph > > > >  Original Message  From: "Andras Balogh" <andras.balogh@...= et> > > To: <gdalgorithmslist@...> > > Sent: Thursday, July 21, 2005 8:02 PM > > Subject: [Algorithms] rotating in world space to have result in screen > > space > > > > > >> I have a point 'v0' in world space, that I can only rotate around a > >> fixed point 'p' and axis 'r'. So movement is constrained on a circle. > >> Now, I'm looking for the point 'v1' on this circle, so that the three > >> points (v0, p, v1), when projected to the screen, enclose a given angl= e. > >> > >> Any ideas how to compute the world space 'v1', given the screen space > >> angle? > >> > >> I was thinking of working backwards, rotating first in screen space, > >> using the given angle, and then trying to figure out the world space > >> position of v1, but I'm a bit afraid, that this wouldn't be a very > >> robust solution.. > >> > >> thanks, > >> > >> > >> Andras >=20 >=20 >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=3D7477&alloc_id=3D16492&op=3Dclic= k > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 > 
From: Andras Balogh <andras.balogh@gm...>  20050721 19:12:17

Ok, so here's the original problem: I have a rectangle in world space, two of its neighbouring vertices are locked at a fixed position (so you can rotate the rectangle around the locked edge). Now, given an arbitrary viewing transform, rotate the rectangle, so that its screen space size is maximized! I was thinking that I'll have the maximum area, if the normal vector of the rotated rectangle is parallel in screen space to the projected locked edge. So, having the screen space normal, I could compute the angle between that and the screen space up vector. If I could transform this angle back to world space, then I could just rotate the rectangle by this amout.. Ehh, I hope it makes some sense, there more I think about it, the more confused I become :) Andras Ralph Egas wrote: > Hi Andras, > > Are you looking for a solution for an arbitrary 'r' and 'p'? What if 'r' > gets perpendicular to the right or up vector in screenspace? Obviously > then there's not going to be any solution for any world space angle.. > > If you could just elaborate your idea and what you need this solution > for, maybe that could help. > > Cheers, > > Ralph > >  Original Message  From: "Andras Balogh" <andras.balogh@...> > To: <gdalgorithmslist@...> > Sent: Thursday, July 21, 2005 8:02 PM > Subject: [Algorithms] rotating in world space to have result in screen > space > > >> I have a point 'v0' in world space, that I can only rotate around a >> fixed point 'p' and axis 'r'. So movement is constrained on a circle. >> Now, I'm looking for the point 'v1' on this circle, so that the three >> points (v0, p, v1), when projected to the screen, enclose a given angle. >> >> Any ideas how to compute the world space 'v1', given the screen space >> angle? >> >> I was thinking of working backwards, rotating first in screen space, >> using the given angle, and then trying to figure out the world space >> position of v1, but I'm a bit afraid, that this wouldn't be a very >> robust solution.. >> >> thanks, >> >> >> Andras 
From: Ralph Egas <ralph.egas@ch...>  20050721 18:41:03

Hi Andras, Are you looking for a solution for an arbitrary 'r' and 'p'? What if 'r' gets perpendicular to the right or up vector in screenspace? Obviously then there's not going to be any solution for any world space angle.. If you could just elaborate your idea and what you need this solution for, maybe that could help. Cheers, Ralph  Original Message  From: "Andras Balogh" <andras.balogh@...> To: <gdalgorithmslist@...> Sent: Thursday, July 21, 2005 8:02 PM Subject: [Algorithms] rotating in world space to have result in screen space >I have a point 'v0' in world space, that I can only rotate around a fixed >point 'p' and axis 'r'. So movement is constrained on a circle. Now, I'm >looking for the point 'v1' on this circle, so that the three points (v0, p, >v1), when projected to the screen, enclose a given angle. > > Any ideas how to compute the world space 'v1', given the screen space > angle? > > I was thinking of working backwards, rotating first in screen space, using > the given angle, and then trying to figure out the world space position of > v1, but I'm a bit afraid, that this wouldn't be a very robust solution.. > > thanks, > > > Andras > > >  > SF.Net email is sponsored by: Discover Easy Linux Migration Strategies > from IBM. Find simple to follow Roadmaps, straightforward articles, > informative Webcasts and more! Get everything you need to get up to > speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click > _______________________________________________ > GDAlgorithmslist mailing list > GDAlgorithmslist@... > https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist > Archives: > http://sourceforge.net/mailarchive/forum.php?forum_id=6188 
From: Andras Balogh <andras.balogh@gm...>  20050721 18:02:32

I have a point 'v0' in world space, that I can only rotate around a fixed point 'p' and axis 'r'. So movement is constrained on a circle. Now, I'm looking for the point 'v1' on this circle, so that the three points (v0, p, v1), when projected to the screen, enclose a given angle. Any ideas how to compute the world space 'v1', given the screen space angle? I was thinking of working backwards, rotating first in screen space, using the given angle, and then trying to figure out the world space position of v1, but I'm a bit afraid, that this wouldn't be a very robust solution.. thanks, Andras 
From: Richard Mitton <rmitton@cl...>  20050721 08:45:09

> hrm is there a metric for how concave a mesh is? At a guess, I'd imagine you could take the ratio between it's surface area, and the surface area of it's convex hull.  ()() Richard Mitton .:. rmitton@... ( '.') (")_(") xbox jester .:. climax action [kingston] _____ From: gdalgorithmslistadmin@... [mailto:gdalgorithmslistadmin@...] On Behalf Of Rowan Wyborn Sent: 20 July 2005 07:48 To: gdalgorithmslist@... Subject: RE: [Algorithms] Hardware friendly mesh subdivision quadtree/octree/KD are good if the mesh is basically convex. bsp tree in this case will perform very poorly. if the mesh has lots of concavities (read internal rooms and hallways), a bsp tree will probably be much more efficient. hrm is there a metric for how concave a mesh is? ie for a sphere it would return 0, for a set of rooms and hallways it would return a high value, and for say a cow it would return something in between? Original Message From: Richard Mitton [mailto:rmitton@...] Sent: Tuesday, 19 July 2005 7:56 PM To: gdalgorithmslist@... Subject: RE: [Algorithms] Hardware friendly mesh subdivision Quadtree, KDtree, BSP tree...... take your pick. I'm a fan of the loose KDtree myself.  ()() Richard Mitton .:. rmitton@... ( '.') (")_(") xbox jester .:. climax action [kingston] _____ From: gdalgorithmslistadmin@... [mailto:gdalgorithmslistadmin@...] On Behalf Of Petr Smilek Sent: 18 July 2005 23:49 To: gdalgorithmslist@... Subject: [Algorithms] Hardware friendly mesh subdivision Hi all ! I have following problem: I have quite large (spatialy) & complex (2050K tris) mesh which is represented by vertex buffer + index buffer + material info table (which maps some kind of material info > range of triangles in index buffer). Mesh is optimized for vcache. I want to implement some kind of 'hardware friendly' spatial subdivision which will allow me efficiently map "some kind of volume description (sphere,AABB,OBB)" > number (smaller the better) of index buffer ranges (preferably still optimized for vcache) representing triangles in specified volume. Ideal subdivision would be hierarchical, but gridlike structure can be OK too. I want to use this structure for obvious reasons  stuff like l visibility culling, lighting (light sphere > portion of mesh affected by light) etc. Subdivision can be done manually by artists of course, which is what I want to avoid. Is there some known solution to such problem ? Thanks Petr 
From: Rowan Wyborn <rowan@ir...>  20050720 06:48:37

quadtree/octree/KD are good if the mesh is basically convex. bsp tree in = this case will perform very poorly. =20 if the mesh has lots of concavities (read internal rooms and hallways), = a bsp tree will probably be much more efficient. =20 hrm is there a metric for how concave a mesh is? ie for a sphere it = would return 0, for a set of rooms and hallways it would return a high = value, and for say a cow it would return something in between? Original Message From: Richard Mitton [mailto:rmitton@...] Sent: Tuesday, 19 July 2005 7:56 PM To: gdalgorithmslist@... Subject: RE: [Algorithms] Hardware friendly mesh subdivision Quadtree, KDtree, BSP tree...... take your pick. =20 I'm a fan of the loose KDtree myself.  ()() Richard Mitton .:. rmitton@... ( '.') (")_(") xbox jester .:. climax action [kingston] =20 _____ =20 From: gdalgorithmslistadmin@... = [mailto:gdalgorithmslistadmin@...] On Behalf Of Petr = Smilek Sent: 18 July 2005 23:49 To: gdalgorithmslist@... Subject: [Algorithms] Hardware friendly mesh subdivision Hi all ! I have following problem:=20 =20 I have quite large (spatialy) & complex (2050K tris) mesh which is = represented by=20 vertex buffer + index buffer + material info table (which maps some kind = of material info > range of triangles in index buffer). Mesh is optimized for vcache.=20 =20 I want to implement some kind of 'hardware friendly' spatial subdivision = which will allow me efficiently map "some kind of volume description (sphere,AABB,OBB)" > number (smaller = the better) of index buffer ranges (preferably still optimized for vcache) representing triangles in specified volume. = Ideal subdivision would be hierarchical, but gridlike structure can be = OK too.=20 I want to use this structure for obvious reasons  stuff like l = visibility culling, lighting (light sphere > portion of mesh affected = by light) etc. Subdivision can be done manually by artists of course, which is what I = want to avoid. =20 Is there some known solution to such problem ? =20 Thanks Petr =20 =20 =20 =20 
From: George Warner <geowar@ap...>  20050719 16:07:13

On Mon, 18 Jul 2005 15:11:53 +0200, Ruslan Shestopalyuk <silver@...> wrote: > There should be an article about that in Game Programming Gems 4, but I > haven't got a chance to read it, unfortunately. FYI: Game Programming Gems 5 is out. ;)  Enjoy, George Warner, Schizophrenic Optimization Scientist Apple Developer Technical Support (DTS) 
From: Andrew Graham <andrew@sp...>  20050719 14:49:04

Is this necessary? Your bsp builder should detect polys that are coplanar to existing partition planes (and facing the same way), and just discard them. At 16:16 19.07.2005, you wrote: >I am working on a bsp tree builder and I wanted to be able to take a >nonuniform triangle mesh and preproccess it into convex ngons before >running it into my bsptree builder. I was accomplishing this by >gathering the normal data and distance D from 0,0,0 along normal to the >intersection of the plane. Then I used that data to put all triangles >into coplanar adjacent triangle groups. Now I need a way to merge the >triangles into the minnimum number of convex polygons. >Thanks for the Help, Kevin > > > >____________________________________________________ >Start your day with Yahoo!  make it your home page >http://www.yahoo.com/r/hs > > > > >SF.Net email is sponsored by: Discover Easy Linux Migration Strategies >from IBM. Find simple to follow Roadmaps, straightforward articles, >informative Webcasts and more! Get everything you need to get up to >speed, fast. http://ads.osdn.com/?ad_id=7477&alloc_id=16492&op=click >_______________________________________________ >GDAlgorithmslist mailing list >GDAlgorithmslist@... >https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist >Archives: >http://sourceforge.net/mailarchive/forum.php?forum_id=6188 